Quick differential geometry questions

  • Thread starter akoska
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  • #1
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Hi,

I just want to know if I'm on the right track with these questions...

1. To prove that a point and a line in R^3 are not surfaces, I showed that the function from an open interval U in R^2 to the intersection of the point/line and a subset W in R^3 cannot be a homeomorphism. This is because the function is not injective.

2. For the curve y= sin(1/x), x does not equal 0, I want to know if the generalized cylinder on this graph is a smooth, regular surface.

Parametrized curve: (u, sin(1/u), 0)

So, sigma(u,v)=(u, sin(1/u), 0) + va

where a is a unit vector in the direction of the translation.

So, I showed that sigma is smooth on R/{0} by showing that all partial derivatives exist at all points except when u=0.

But, I'm having trouble showing the 'regular' part.

sigma_u = (1, -1/u^2 (cos(1/u)), 0)

sigma_v=a

So, for the surface to be regular, the cross product of the above two functions cannot be zero. So, the surface is regular if the tangent vector of the parametrized curve is not parallel to a.

I'm having trouble with this problem because by how the question is phrased, it seems like I'm only supposed to find a yes/no answer, not a 'depending on the vector a' answer. Do you see what I mean? So, I don't know if I'm missing anything, not proving rigorously enough... etc.

Any help would be greatly appreciated!
 

Answers and Replies

  • #2
Dick
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I'm a little unclear as to what you did with 1) to show there is no homeomorphism. But for part 2), as you point out there are many generalized cylinders over the curve. Since the question says 'THE generalized cylinder', I think I would assume they meant a=unit vector in the z-direction.
 
  • #3
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Ok, thanks.

For 1) I have to show that the function is not a homeomorphism (bijective, function and inverse of the function is continuous). So, I showed that the function is not injective:

For a point: the intersection of the point and a subset W is just a point, p. So the function going from an interval to a point (a constant value) is not injective.

For a line: If u is a unit vector, y=au (a is a scalar) describes a line pointing in direction of u. So, y is a function of 1 variable. So the function going from an interval (of 2 variables) to 1 variable is not injective.

Is this rigorous enough?
 
  • #4
Dick
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For a point you are on pretty safe ground. For a line, what is the theorem that says a mapping from a higher dimension space to a lower dimension space can't be injective? Offhand, I don't remember. Can you remind me?
 
  • #5
Dick
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Ah, it's because invariance of dimension actually depends on proofs from algebraic topology. But in the case of mapping two dimensions to one its pretty easy. A line is split in two by removing a point but a two dimensional surface is not. But this property is a homeomorphism invariant. If none of this makes any sense, then just go ahead and say it's obviously true. Because it is.
 

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